01 Monotonicity and constrained extrema Use derivative sign changes, endpoints and feasible domains to distinguish local extrema from global optima. ↗ Question directory1. Foundation · Locate and classify the stationary points of x³−3x. 2. Standard · Find the global maximum of x³−3x on [−2,1]. 3. Transfer · A rectangle has perimeter 20. Find its maximum area and justify t 4. Foundation · Where is x² decreasing? 5. Foundation · Is zero an extremum of x³? 6. Foundation · Find the minimum of x² on [1,3]. 7. Foundation · For f=x²−4x, find the local minimum value. 8. Standard · Explain how |x| can have a minimum at zero despite no derivative 9. Standard · If f′=(x−1)²(x+2), classify x=−2 and x=1. 10. Standard · Does the local maximum of x³−3x give a global maximum on all real 11. Standard · For f=x⁴, does f″(0)=0 prevent zero being a minimum? 12. Transfer · For f=x³−kx, determine for which real k it has two distinct local 13. Transfer · Find the supremum of 2x−x² on (0,1). Is it attained? ← →
What you will be able to explain Use derivative sign changes, endpoints and feasible domains to distinguish local extrema from global optima. Handle parameter boundaries and non-attained bounds. Candidates need classification Optimization is a comparison over an allowed set, not just solving an equation.
A derivative root suggests a candidate; nearby signs explain whether direction reverses.
f ′ < 0 → f ′ > 0 : local minimum f'<0\to f'>0:\ \text{local minimum} f ′ < 0 → f ′ > 0 : local minimum A derivative root suggests a candidate; nearby signs explain whether direction reverses.
f = x 3 − k x ; f ′ = 3 x 2 − k f=x^3-kx;\ f'=3x^2-k f = x 3 − k x ; f ′ = 3 x 2 − k A derivative root suggests a candidate; nearby signs explain whether direction reverses.
k > 0 : x = ± k / 3 ; signs + , − , + k>0:\ x=\pm\sqrt{k/3};\ \text{signs }+,-,+ k > 0 : x = ± k /3 ; signs + , − , + When it applies Closed intervals require both endpoints; open-domain bounds may not be attained. Counterexample: x³ has derivative zero at zero but keeps increasing; |x| has a minimum at a corner.
Sign chart, graph, endpoint-value table and feasible-set inequality. Pause at these common mistakes Calling every stationary point an extremum. Reporting a limit at an excluded endpoint as an attained maximum. PREDICT → EXPLORE → EXPLAIN → TRANSFER
Make a prediction, then explore the relationship. Lesson question: Predict the stationary points of x³−kx for k<0, k=0 and k>0.
Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.
Numerical data for the diagram; printable Candidate input Function value Role -2 -2 Endpoint / 端點 -1 2 Local maximum / 局部極大 1 -2 Endpoint / 端點
Explain: Compare the function with derivative 3x²−k and inspect the signs at the roots.
Transfer: Change the visible interval and explain why a global optimum also needs endpoint comparisons.
My prediction, exploration and explanation Try it. Leave your reasoning visible. Use one hint at a time. A correction explains what changed, not just the final answer.
Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.
Locate and classify the stationary points of x³−3x. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Classify using the derivative signs on adjacent intervals.
Hint 2+ Start from this intermediate relation.
f ′ = 3 ( x − 1 ) ( x + 1 ) f'=3(x-1)(x+1) f ′ = 3 ( x − 1 ) ( x + 1 ) Worked solution+ Classify using the derivative signs on adjacent intervals.
Establish and simplify this mathematical relation.
f ′ = 3 ( x − 1 ) ( x + 1 ) f'=3(x-1)(x+1) f ′ = 3 ( x − 1 ) ( x + 1 ) Establish and simplify this mathematical relation.
signs: + , − , + ; f ( − 1 ) = 2 ; f ( 1 ) = − 2 \text{signs: }+,-,+;\ f(-1)=2;\ f(1)=-2 signs: + , − , + ; f ( − 1 ) = 2 ; f ( 1 ) = − 2 Both roots require a sign-change test.
Local maximum (−1,2); local minimum (1,−2).
Checks and common pitfalls: Both roots require a sign-change test.
Reasoning checklist · self / teacher assessment Classify using the derivative signs on adjacent intervals. Give the intermediate relations and explain the conclusion with its restrictions. Both roots require a sign-change test. Think first. Reveal a hint when the class is ready.
Find the global maximum of x³−3x on [−2,1]. f = x 3 − 3 x ; − 2 ≤ x ≤ 1 f=x^3-3x;\ -2\le x\le1 f = x 3 − 3 x ; − 2 ≤ x ≤ 1 First step provided — complete the reasoning: Compare endpoints and interior stationary values.
Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Compare endpoints and interior stationary values.
Hint 2+ Start from this intermediate relation.
f ( − 2 ) = − 2 ; f ( − 1 ) = 2 ; f ( 1 ) = − 2 f(-2)=-2;\ f(-1)=2;\ f(1)=-2 f ( − 2 ) = − 2 ; f ( − 1 ) = 2 ; f ( 1 ) = − 2 Worked solution+ Compare endpoints and interior stationary values.
Establish and simplify this mathematical relation.
f ( − 2 ) = − 2 ; f ( − 1 ) = 2 ; f ( 1 ) = − 2 f(-2)=-2;\ f(-1)=2;\ f(1)=-2 f ( − 2 ) = − 2 ; f ( − 1 ) = 2 ; f ( 1 ) = − 2 Establish and simplify this mathematical relation.
Stationary points alone do not determine a closed-interval maximum.
The requested value is 2.
Checks and common pitfalls: Stationary points alone do not determine a closed-interval maximum.
Think first. Reveal a hint when the class is ready.
A rectangle has perimeter 20. Find its maximum area and justify the feasible domain. 2 x + 2 y = 20 ; x , y > 0 2x+2y=20;\ x,y>0 2 x + 2 y = 20 ; x , y > 0 Both rectangle lengths are positive; degenerate endpoints are excluded. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Eliminate one length and optimize only positive rectangles.
Hint 2+ Start from this intermediate relation.
A = x ( 10 − x ) ; 0 < x < 10 ; A ′ = 10 − 2 x A=x(10-x);\ 0<x<10;\ A'=10-2x A = x ( 10 − x ) ; 0 < x < 10 ; A ′ = 10 − 2 x Worked solution+ Eliminate one length and optimize only positive rectangles.
Establish and simplify this mathematical relation.
A = x ( 10 − x ) ; 0 < x < 10 ; A ′ = 10 − 2 x A=x(10-x);\ 0<x<10;\ A'=10-2x A = x ( 10 − x ) ; 0 < x < 10 ; A ′ = 10 − 2 x Establish and simplify this mathematical relation.
x = 5 ; y = 5 ; A = 25 x=5;\ y=5;\ A=25 x = 5 ; y = 5 ; A = 25 The endpoints are degenerate rectangles and are excluded, but the interior optimum is attainable.
The requested value is 25.
Checks and common pitfalls: The endpoints are degenerate rectangles and are excluded, but the interior optimum is attainable.
Think first. Reveal a hint when the class is ready.
Where is x² decreasing? Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Use the derivative sign rather than whether x² is positive.
Hint 2+ Start from this intermediate relation.
Worked solution+ Use the derivative sign rather than whether x² is positive.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
2 x < 0 ⟺ x < 0 2x<0\iff x<0 2 x < 0 ⟺ x < 0 A positive function can decrease.
On (−∞,0).
Checks and common pitfalls: A positive function can decrease.
Reasoning checklist · self / teacher assessment Use the derivative sign rather than whether x² is positive. Give the intermediate relations and explain the conclusion with its restrictions. A positive function can decrease. Think first. Reveal a hint when the class is ready.
Is zero an extremum of x³? Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Check signs on both sides of the stationary point.
Hint 2+ Start from this intermediate relation.
Worked solution+ Check signs on both sides of the stationary point.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
f ′ ( x ) = 3 x 2 > 0 ( x ≠ 0 ) f'(x)=3x^2>0\ (x\ne0) f ′ ( x ) = 3 x 2 > 0 ( x = 0 ) A zero derivative is necessary for an interior differentiable extremum, not sufficient.
No; f′=3x² is positive on both sides.
Checks and common pitfalls: A zero derivative is necessary for an interior differentiable extremum, not sufficient.
Reasoning checklist · self / teacher assessment Check signs on both sides of the stationary point. Give the intermediate relations and explain the conclusion with its restrictions. A zero derivative is necessary for an interior differentiable extremum, not sufficient. Think first. Reveal a hint when the class is ready.
Find the minimum of x² on [1,3]. f = x 2 ; 1 ≤ x ≤ 3 f=x^2;\ 1\le x\le3 f = x 2 ; 1 ≤ x ≤ 3 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Recognize the stationary point lies outside the interval.
Hint 2+ Start from this intermediate relation.
f ′ = 2 x > 0 ( 1 ≤ x ≤ 3 ) f'=2x>0\ (1\le x\le3) f ′ = 2 x > 0 ( 1 ≤ x ≤ 3 ) Worked solution+ Recognize the stationary point lies outside the interval.
Establish and simplify this mathematical relation.
f ′ = 2 x > 0 ( 1 ≤ x ≤ 3 ) f'=2x>0\ (1\le x\le3) f ′ = 2 x > 0 ( 1 ≤ x ≤ 3 ) Establish and simplify this mathematical relation.
min f = f ( 1 ) = 1 \min f=f(1)=1 min f = f ( 1 ) = 1 An endpoint minimum need not have zero derivative.
The requested value is 1.
Checks and common pitfalls: An endpoint minimum need not have zero derivative.
Think first. Reveal a hint when the class is ready.
For f=x²−4x, find the local minimum value. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Solve the first derivative and confirm positive curvature.
Hint 2+ Start from this intermediate relation.
f ′ = 2 x − 4 = 0 ⟹ x = 2 ; f ′ ′ = 2 f'=2x-4=0\implies x=2;\ f''=2 f ′ = 2 x − 4 = 0 ⟹ x = 2 ; f ′′ = 2 Worked solution+ Solve the first derivative and confirm positive curvature.
Establish and simplify this mathematical relation.
f ′ = 2 x − 4 = 0 ⟹ x = 2 ; f ′ ′ = 2 f'=2x-4=0\implies x=2;\ f''=2 f ′ = 2 x − 4 = 0 ⟹ x = 2 ; f ′′ = 2 Establish and simplify this mathematical relation.
The point’s x-coordinate 2 differs from its minimum value −4.
The requested value is -4.
Checks and common pitfalls: The point’s x-coordinate 2 differs from its minimum value −4.
Think first. Reveal a hint when the class is ready.
Explain how |x| can have a minimum at zero despite no derivative there. A derivative requires equal finite one-sided limits; an extremum itself does not require differentiability. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Use the definition before applying a differentiability-based test.
Hint 2+ Start from this intermediate relation.
∣ x ∣ ≥ 0 = ∣ 0 ∣ |x|\ge0=|0| ∣ x ∣ ≥ 0 = ∣0∣ Worked solution+ Use the definition before applying a differentiability-based test.
Establish and simplify this mathematical relation.
∣ x ∣ ≥ 0 = ∣ 0 ∣ |x|\ge0=|0| ∣ x ∣ ≥ 0 = ∣0∣ Establish and simplify this mathematical relation.
f ′ ( 0 ) does not exist f\prime(0)\text{ does not exist} f ′ ( 0 ) does not exist The stationary-point test does not cover corners.
|x|≥0 with equality at zero; differentiability is not required by the definition of a minimum.
Checks and common pitfalls: The stationary-point test does not cover corners.
Reasoning checklist · self / teacher assessment Use the definition before applying a differentiability-based test. Give the intermediate relations and explain the conclusion with its restrictions. The stationary-point test does not cover corners. Think first. Reveal a hint when the class is ready.
If f′=(x−1)²(x+2), classify x=−2 and x=1. f ′ = ( x − 1 ) 2 ( x + 2 ) f'=(x-1)^2(x+2) f ′ = ( x − 1 ) 2 ( x + 2 ) Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Even-multiplicity factors do not change sign.
Hint 2+ Start from this intermediate relation.
sign ( f ′ ) = sign ( x + 2 ) ( x ≠ 1 ) \text{sign}(f')=\text{sign}(x+2)\ (x\ne1) sign ( f ′ ) = sign ( x + 2 ) ( x = 1 ) Worked solution+ Even-multiplicity factors do not change sign.
Establish and simplify this mathematical relation.
sign ( f ′ ) = sign ( x + 2 ) ( x ≠ 1 ) \text{sign}(f')=\text{sign}(x+2)\ (x\ne1) sign ( f ′ ) = sign ( x + 2 ) ( x = 1 ) Establish and simplify this mathematical relation.
at − 2 : − → + ; at 1 : + → + \text{at }-2:-\to+;\ \text{at }1:+\to+ at − 2 : − → + ; at 1 : + → + A repeated derivative root need not produce an extremum.
−2 is a local minimum; 1 is stationary but not an extremum.
Checks and common pitfalls: A repeated derivative root need not produce an extremum.
Reasoning checklist · self / teacher assessment Even-multiplicity factors do not change sign. Give the intermediate relations and explain the conclusion with its restrictions. A repeated derivative root need not produce an extremum. Think first. Reveal a hint when the class is ready.
Does the local maximum of x³−3x give a global maximum on all real numbers? Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Compare the local behavior with unbounded-domain behavior.
Hint 2+ Start from this intermediate relation.
Worked solution+ Compare the local behavior with unbounded-domain behavior.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
lim x → + ∞ ( x 3 − 3 x ) = + ∞ \lim_{x\to+\infty}(x^3-3x)=+\infty x → + ∞ lim ( x 3 − 3 x ) = + ∞ Local extremum and global optimum are different claims.
No: f(x)→∞ as x→∞.
Checks and common pitfalls: Local extremum and global optimum are different claims.
Reasoning checklist · self / teacher assessment Compare the local behavior with unbounded-domain behavior. Give the intermediate relations and explain the conclusion with its restrictions. Local extremum and global optimum are different claims. Think first. Reveal a hint when the class is ready.
For f=x⁴, does f″(0)=0 prevent zero being a minimum? Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ When the second-derivative test is inconclusive, return to signs or the definition.
Hint 2+ Start from this intermediate relation.
f ′ ′ ( 0 ) = 0 ; f ′ = 4 x 3 f''(0)=0;\ f'=4x^3 f ′′ ( 0 ) = 0 ; f ′ = 4 x 3 Worked solution+ When the second-derivative test is inconclusive, return to signs or the definition.
Establish and simplify this mathematical relation.
f ′ ′ ( 0 ) = 0 ; f ′ = 4 x 3 f''(0)=0;\ f'=4x^3 f ′′ ( 0 ) = 0 ; f ′ = 4 x 3 Establish and simplify this mathematical relation.
x 4 ≥ 0 = f ( 0 ) x^4\ge0=f(0) x 4 ≥ 0 = f ( 0 ) Zero second derivative means inconclusive, not no extremum.
No; x⁴≥0, and f′ changes from negative to positive.
Checks and common pitfalls: Zero second derivative means inconclusive, not no extremum.
Reasoning checklist · self / teacher assessment When the second-derivative test is inconclusive, return to signs or the definition. Give the intermediate relations and explain the conclusion with its restrictions. Zero second derivative means inconclusive, not no extremum. Think first. Reveal a hint when the class is ready.
For f=x³−kx, determine for which real k it has two distinct local extrema. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Separate the number of derivative roots from their sign changes.
Hint 2+ Start from this intermediate relation.
f ′ = 3 x 2 − k = 0 ⟹ x 2 = k / 3 f'=3x^2-k=0\implies x^2=k/3 f ′ = 3 x 2 − k = 0 ⟹ x 2 = k /3 Worked solution+ Separate the number of derivative roots from their sign changes.
Establish and simplify this mathematical relation.
f ′ = 3 x 2 − k = 0 ⟹ x 2 = k / 3 f'=3x^2-k=0\implies x^2=k/3 f ′ = 3 x 2 − k = 0 ⟹ x 2 = k /3 Establish and simplify this mathematical relation.
k > 0 : + , − , + ; k = 0 : + , 0 , + ; k < 0 : f ′ > 0 k>0:\ +,-,+;\ k=0:\ +,0,+;\ k<0:\ f\prime>0 k > 0 : + , − , + ; k = 0 : + , 0 , + ; k < 0 : f ′ > 0 Boundary k=0 does not belong to the two-extrema regime.
Exactly k>0. At k=0 there is a stationary non-extremum; at k<0 the function increases everywhere.
Checks and common pitfalls: Boundary k=0 does not belong to the two-extrema regime.
Reasoning checklist · self / teacher assessment Separate the number of derivative roots from their sign changes. Give the intermediate relations and explain the conclusion with its restrictions. Boundary k=0 does not belong to the two-extrema regime. Think first. Reveal a hint when the class is ready.
Find the supremum of 2x−x² on (0,1). Is it attained? f = 2 x − x 2 ; 0 < x < 1 f=2x-x^2;\ 0<x<1 f = 2 x − x 2 ; 0 < x < 1 The allowed interval is open; its endpoint limit is not an attained value. Skills and prerequisite lessons Applications of derivatives Monotonicity and constrained extrema Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Use monotonicity and the excluded endpoint limit.
Hint 2+ Start from this intermediate relation.
f ′ = 2 − 2 x > 0 ( 0 < x < 1 ) f'=2-2x>0\ (0<x<1) f ′ = 2 − 2 x > 0 ( 0 < x < 1 ) Worked solution+ Use monotonicity and the excluded endpoint limit.
Establish and simplify this mathematical relation.
f ′ = 2 − 2 x > 0 ( 0 < x < 1 ) f'=2-2x>0\ (0<x<1) f ′ = 2 − 2 x > 0 ( 0 < x < 1 ) Establish and simplify this mathematical relation.
f ( x ) = 1 − ( x − 1 ) 2 < 1 ; lim x → 1 − f ( x ) = 1 f(x)=1-(x-1)^2<1;\ \lim_{x\to1^-}f(x)=1 f ( x ) = 1 − ( x − 1 ) 2 < 1 ; x → 1 − lim f ( x ) = 1 An excluded endpoint limit is not an attained maximum.
Supremum 1, no maximum attained.
Checks and common pitfalls: An excluded endpoint limit is not an attained maximum.
Reasoning checklist · self / teacher assessment Use monotonicity and the excluded endpoint limit. Give the intermediate relations and explain the conclusion with its restrictions. An excluded endpoint limit is not an attained maximum. Think first. Reveal a hint when the class is ready.
